Exact sequences on Powell–Sabin splits

Abstract

We construct smooth finite elements spaces on Powell–Sabin triangulations that form an exact sequence. The first space of the sequence coincides with the classical \(C^1\) Powell–Sabin space, while the others form stable and divergence-free yielding pairs for the Stokes problem. We develop degrees of freedom for these spaces that induce projections that commute with the differential operators.

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Acknowledgements

J. Guzman and A. Lischke were supported by the NSF grant DMS-1913083. M. Neilan was supported by the NSF grant DMS-1719829.

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Appendix

Appendix

Proof of Lemma 7

Proof

Suppose \(z \in W_r(\{a,m,b\})\) is such that (4.1a)–(4.1c) are all zero. We will show that z must be identically zero on [ab]. Let \(\psi (x)\) be a degree r polynomial on the interval [0, 1] satisfying

$$\begin{aligned} \begin{aligned} \psi (0)&= 1, \quad \psi (1) = 0, \\ \int _0^1 \psi (x) p(x)&= 0 \quad \forall p \in \mathcal {P}_{r-2}([0,1]). \end{aligned} \end{aligned}$$
(A.1)

We note that these conditions uniquely determine \(\psi\). Since z is continuous at m and equal to zero at a and b, and in view of (4.1b)–(4.1c), it follows that z may be represented by

$$\begin{aligned} z(y)&= z(m){\left\{ \begin{array}{ll} \psi \left( \frac{y-m}{a-m}\right) &{} y \in [a,m], \\ \psi \left( \frac{m-y}{m-b}\right) &{} y \in [m,b]. \end{array}\right. } \end{aligned}$$

Since \(z'(y)\) is continuous at m, it must hold that

$$\begin{aligned} \frac{-1}{m-b} \psi '(0) = \frac{1}{a-m}\psi '(0). \end{aligned}$$

Furthermore, given the conditions (A.1) on \(\psi\), we can show that \(\psi '(0) \ne 0\). Suppose that \(\psi '(0) = 0\) in addition to (A.1). Then for any \(p \in \mathcal {P}_{r-1}([0,1])\) with \(p(0) = 0\),

$$\begin{aligned} \int _0^1 \psi '(x) p(x)&= -\int _0^1 \psi (x) p'(x) + \psi (1) p(1) - \psi (0) p(0) = -\int _0^1 \psi (x) p'(x) = 0 \end{aligned}$$

since \(p'(x) \in \mathcal {P}_{r-2}([0,1])\). But \(\psi '(x)\) is itself such a function p(x), so it follows that

$$\begin{aligned} \int _0^1 |\psi '(x)|^2 = 0. \end{aligned}$$

Then \(\psi '(x) = 0\), and \(\psi\) is constant on [0, 1]. This contradicts (A.1), so \(\psi '(0) \ne 0\). Furthermore, since \(1/(b-m) \ne 1/(a-m)\), it follows that \(z(m) = 0\). Therefore \(z = 0\) on [ab]. \(\square\)

Proof of Theorem 3

Proof

(1) Proof of (4.10a). Let \(p \in C^\infty (T)\) and \(\rho := \text {rot }\varPi _0^r p - \varpi _1^{r-1} \text {rot }p \in S_{r-1}^1(T^{\mathrm{ps}})\). We show that \(\rho\) vanishes on (4.5).

First,

$$\begin{aligned} \rho (z_i)&= \text {rot }\varPi _0^r p(z_i) - \varpi _1^{r-1} \text {rot }p(z_i) = 0,\\ \mathrm{div}\,\rho (z_i)&= -\mathrm{div}\,\varphi _1^{r-1} \text {rot }p(z_i) = -\mathrm{div}\,\text {rot }p(z_i) = 0, \end{aligned}$$

by the definitions of \(\varPi _0^r\) and \(\varpi _1^{r-1}\) along with DOFs (4.2a) and (4.5a).

Next, if \(r=2\),

$$\begin{aligned} \int _{e_i} \rho \cdot n_i&= \int _{e_i} \big (\text {rot }\varPi _0^r p - \varpi _1^{r-1} \text {rot }p \big )\cdot n_i\\&= \int _{e_i} \big (\text {rot }\varPi _0^r p - \varPi _1^{r-1} \text {rot }p \big )\cdot n_i=0, \end{aligned}$$

using (4.5b), (4.3b) and (4.7b). Similar arguments show that, for \(r\ge 3\),

$$\begin{aligned} \rho (z_{3+i})\cdot n_i&= (\text {rot }\varPi _0^r p (z_{3+i}) - \varPi _1^{r-1} \text {rot }p(z_{3+i}))\cdot n_i =0,\\ \int _e \rho \cdot w&= \int _e (\text {rot }\varPi _0^r p - \varpi _1^{r-1} \text {rot }p)\cdot w = \int _e (\text {rot }\varPi _0^r p - \varPi _1^{r-1} \text {rot }p)\cdot w = 0, \end{aligned}$$

and

$$\begin{aligned} \int _T \rho \cdot \text {rot }w&= \int _T (\text {rot }\varPi _0^r p - \varPi _1^{r-1} \text {rot }p) \cdot w = 0. \end{aligned}$$

Next using (4.5c) gives

$$\begin{aligned} \mathrm{div}\,\rho (z_{3+i}) = -\mathrm{div}\,\varpi _1^{r-1} \text {rot }p (z_{3+i}) = -\mathrm{div}\,\text {rot }p (z_{3+i}) = 0, \end{aligned}$$

and (4.5e) yields

$$\begin{aligned} \int _e (\mathrm{div}\,\rho ) q = -\int _e (\mathrm{div}\,\varpi _1^{r-1} \text {rot }p) q = -\int _e (\mathrm{div}\,\text {rot }p)q = 0 \end{aligned}$$

for all \(q\in \mathcal {P}_{r-4}(e)\) and \(e\in {\mathcal {E}}^b(T^{\mathrm{ps}})\). The same arguments, but using (4.5g), gives

$$\begin{aligned} \int _T (\mathrm{div}\,\rho ) q=0\qquad \forall q\in \mathring{L}_{r-1}^2(T^{\mathrm{ps}}). \end{aligned}$$

Applying Lemma 12 shows that \(\rho \equiv 0\), and so (4.10a) holds.

(2) Proof of (4.10b). For some \(v \in [C^\infty (T)]^2\), we define \(\rho := {\mathop {\mathrm {div}\,}}\varpi _1^{r-1} v - \varpi _2^{r-2} {\mathop {\mathrm {div}\,}}v \in L_{r-2}^2(T^{\mathrm{ps}})\). Then we need only show that \(\rho\) is zero for all DOFs in (4.6). For the vertex DOFs, we have for each \(z_i\),

$$\begin{aligned} \rho (z_i)&= {\mathop {\mathrm {div}\,}}\varpi _1^{r-1} v(z_i) - \varpi _2^{r-2} {\mathop {\mathrm {div}\,}}v(z_i) = 0, \end{aligned}$$

by (4.5a) and (4.6a). Next, for each \(i = 1,2,3\),

$$\begin{aligned} \rho (z_{3+i})&= {\mathop {\mathrm {div}\,}}\varpi _1^{r-1} v(z_{3+i}) - \varpi _2^{r-2} {\mathop {\mathrm {div}\,}}v(z_{3+i}) = 0, \end{aligned}$$

where we have used (4.5a) and (4.6b). Similar arguments show that

$$\begin{aligned} \int _e \rho q =0\qquad \forall q\in \mathcal {P}_{r-4}(e),\ e\in {\mathcal {E}}^b(T^{\mathrm{ps}}), \end{aligned}$$

by (4.5e) and (4.6c), and that

$$\begin{aligned} \int _T \rho q = 0\qquad \forall q\in \mathring{L}_{r-2}^2(T^{\mathrm{ps}}) \end{aligned}$$

by (4.5g) and (4.6e). Using (4.6d) and (4.5b) if \(r = 2\) or (4.5d) if \(r > 2\),

$$\begin{aligned} \int _T \rho&= \int _T {\mathop {\mathrm {div}\,}}\varpi _1^{r-1} v -\varpi _2^{r-2} {\mathop {\mathrm {div}\,}}v = \int _T {\mathop {\mathrm {div}\,}}(\varpi _1^{r-1} v - v) = \int _{{\partial }T} (\varpi _1^{r-1} v - v )\cdot n = 0. \end{aligned}$$

Therefore, \(\rho \equiv 0\) on T by Lemma 13, and (4.10b) is proved. \(\square\)

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Guzmán, J., Lischke, A. & Neilan, M. Exact sequences on Powell–Sabin splits. Calcolo 57, 13 (2020). https://doi.org/10.1007/s10092-020-00361-x

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Keywords

  • Finite elements
  • Exact sequences
  • Commuting diagrams
  • Powell–Sabin triangulations

Mathematics Subject Classification

  • 65N30